How to find $\lim_{x \to \infty} \arctan(e^x)$? I am posting this with my own solution, to try and remember this concept, which I just spent a long time trying to understand. If I've got anything wrong, please let me know.
Essentially, I am trying to figure out how to find 
$$\lim_{x \to \infty} \arctan(e^x)$$
How can I solve this limit?
 A: Just use substitutio: set $t=\mathrm e^x$.


*

*If $x\to+\infty$, $t\to +\infty$, so $\;\lim_{x\to +\infty}\arctan\mathrm e^x=\lim_{t\to +\infty}\arctan t=\dfrac\pi2$.

*If $x\to-\infty$, $t\to 0$, so $\;\lim_{x\to -\infty}\arctan\mathrm e^x=\lim_{t\to 0}\arctan t=0$.

A: This problem can be solved by dividing it into smaller parts.
Let $f(x) = e^x$ and $g(x) = \arctan(x)$.
I know the range of the function $\arctan(e^x)$ is equal to the intersection of the ranges of $f(x)$ and $g(x)$.
Since the range of $f(x)$ is $[0, \infty)$ and the range of $g(x)$ is $[\frac{-\pi}{2}, \frac{\pi}{2}]$, the intersection is $[0, \infty) \cap [\frac{-\pi}{2}, \frac{\pi}{2}]$ or $[0, \frac{\pi}{2}]$
However, I am not looking for range, but the limit as $x$ approaches infinity, or $\lim_{x \to \infty}$. 
Since the highest number in the possible range is $\frac{\pi}{2}$, that is the limit as $x$ approaches infinity.
if x is approaching negative infinity, or $\lim_{x \to -\infty}$, the answer would be the lowest value in the function's range. Therefore, it would be $0$.
A: It's simple. As $x\rightarrow\infty$, $e^x\rightarrow\infty$. Just think of its graph. As you go to the right, the $e^x$ curve just shoots up into positive infinity. Therefore, the argument of the inverse tangent function is a number that's going to positive infinity. What value does the inverse tangent function approach as its argument tends to positive infinity? Well, it's $\frac{\pi}{2}$, of course. Again, you can see that visually if you take a look at the graph of $f(x)=\arctan{x}$:

